D'Alembert, Lagrange, and Reduction of Order - The Modern Method

Author(s):
Sarah Cummings (Wittenberg University) and Adam E. Parker (Wittenberg University)

The modern technique for reduction of order proceeds as follows.  Given a linear differential equation

$a_m(x)y^{(m)} + a_{m-1}(x) y^{(m-1)} + \dots + a_1(x) y'+a_0(x) y = f(x)$

and a solution $$y_1$$ to the associated homogenous equation

$a_m(x) y^{(m)} + a_{m-1}(x) y^{(m-1)} + \dots + a_1(x) y'+a_0(x) y = 0,$

suppose that a solution to the non-homogeneous equation has the form $$y_2=u y_1$$.  Here, $$u$$ is an unknown function of $$x$$.  We substitute $$y_2$$ along with all its derivatives into the first equation.  After carrying out the product rule, every $$k$$th derivative $$(y_2)^{(k)}=(u y_1)^{(k)}$$ will have a term of the form $$u(y_1^{(k)})$$ and these are the only terms containing $$u$$.  After substituting into the first equation, we collect these $$u$$ terms to find

$a_m(x) (u y_1)^{(m)} + a_{m-1}(x)(u y_1)^{(m-1)} + \dots + a_1(x) (u y_1)'+a_0(x) (u y_1) =$

$A(x, y_1, y_1', \dots, y_1^{(m-1)}, u', u'', \dots, u^{(m)}) +$

$u\,[a_m(x) (y_1^{m}) + a_{m-1}(x) (y_1^{(m-1)}) + \dots + a_1(x) (y_1')+a_0(x) y_1] = f(x),$

where $$A$$ is a linear differential equation in the unknown function $$u$$.  Since $$y_1$$ solves the second equation above, the bracketed expression vanishes.  Hence the first equation above is reduced to

$A(x, y_1, y_1', \dots, y_1^{(m-1)}, u', u'', \dots, u^{(m)}) =f(x).$

As there are no $$u$$ terms in this expression, a change of variables of $$w=u'$$ gives a linear differential equation for the unknown function $$w$$ of order $$m-1$$; namely,

$A(x, y_1, y_1', \dots, y_1^{(m-1)}, w, w', \dots, w^{(m-1)}) = f(x).$

This shows we can reduce the order of a linear differential equation if one solution of the associated homogeneous equation is known.  While not often taught, it is also true that if $$L_m(y)=0$$ is an $$m$$th order equation with known solutions $$y_1, y_2, \dots, y_k,$$ we can reduce the order to $$m-k$$.  Unfortunately, the obvious idea of using all known solutions to successively reduce $$L_{m}(y)=0$$ won't work since typically none of $$y_2,\dots, y_k$$ solve $$L_{m-1}(y)=0$$.

The correct process is described by Ince in [9, p. 121].  Suppose that $$k$$ solutions $$y_1, \dots, y_k$$ are known to $$L_m(y)=0$$.  Use $$v_1=y_1$$ to reduce $$L_m(y)=0$$ to $$L_{m-1}(y)=0$$.  As we mentioned, $$y_2$$ isn't generally a solution to $$L_{m-1}(y)=0$$.  However, $$v_2=\left(\frac{y_2}{v_1}\right)'=\left(\frac{y_2}{y_1}\right)'$$ is a solution, and it can be used to reduce $$L_{m-1}(y)=0$$ to $$L_{m-2}(y)=0$$.  Then $$v_3=\left(\frac{1}{v_2}\left(\frac{y_3}{y_1}\right)'\right)'$$  is a solution to $$L_{m-2}(y)=0$$, and it can be used to reduce $$L_{m-2}(y)=0$$ to $$L_{m-3}(y)=0$$.  Then $$v_4=\left(\frac{1}{v_3}\left(\frac{1}{v_2}\left(\frac{y_4}{y_1}\right)'\right)'\right)'$$ is a solution to $$L_{m-3}(y)=0$$, and it can be used to  reduce that equation.  The process continues with as many solutions as were given at the start.